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A daredevil plans to bungee-jump from a hot-air balloon 65 meters above a carnival midway. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 10 meters above the ground. Model his body as a particle and the cord as having negligible mass and obeying hooke's law. In a preliminary test, hanging at rest from a 5 meter length of the cord, the daredevil finds his body weight stretches the cord by 1.5 meters. He intends to drop from rest at the point where the top end of a longer section of the cord is attached to the stationary hot-air balloon. A) What lenght of cord should he use? B) What maximum acceleration will he experience?

Respuesta :

Answer:

Part a)

[tex]L = 25.81 m[/tex]

Part b)

[tex]a = 27.14 m/s^2[/tex]

Explanation:

Let the length of the string is L

so here if the displacement from initial length is "x" then we have

[tex]L + x = 65 - 10[/tex]

[tex]x = 55 - L[/tex]

now by energy conservation we have

[tex]\frac{1}{2}kx^2 = mgh[/tex]

[tex]\frac{1}{2}k(55 - L)^2 = mg(55)[/tex]

now we also know that for 5 m length of the chord

[tex]mg = kx[/tex]

[tex]k = \frac{mg}{1.5}[/tex]

so for L length of the chord we have

[tex]k = \frac{mg}{1.5} \times \frac{5}{L}[/tex]

now from above equation

[tex]\frac{1}{2}\times \frac{5mg}{1.5 L} (55 - L)^2 = 55 mg[/tex]

[tex]5(55 - L)^2 = 55\times 2\times 1.5L[/tex]

so by solving above we have

[tex]L = 25.81 m[/tex]

Part b)

Now when string is stretched by maximum amount then the acceleration will be maximum

so it is given as

[tex]F_s - mg = ma[/tex]

[tex]k(55 - L) - mg = ma[/tex]

[tex]\frac{5mg}{1.5L} (55 - L) - mg = ma[/tex]

[tex]a = \frac{5g}{1.5(25.81)}(55 - 25.81) - g[/tex]

[tex]a = 27.14 m/s^2[/tex]

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